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Calculus with polynomials

Polynomials are perhaps the simplest functions to do calculus with. Their derivatives and integrals are given by the following rules:

Hence the derivative of x¹⁰⁰ is 100x⁹⁹ and the integral of x¹⁰⁰ is x¹⁰¹/101 + c.

Proof

Because differentiation is linear, we have:

<math>\frac{d\left( \sum_{r=0}^n a_r x^r \right)}{dx} =

\sum_{r=0}^n \frac{d\left(a_r x^r\right)}{dx} = \sum_{r=0}^n a_r \frac{d\left(x^r\right)}{dx}</math>

So it remains to find <math>\frac{d\left(x^r\right)}{dx}</math> for any natural number r. The derivative of function f(x) is given by Newton's difference quotient

<math> f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} </math>

By the binomial theorem, and using the C-notation of combinations,

<math>\left(x+h\right)^r = \sum_{k=0}^r {}^rC_k h^k x^{r-k}</math>

and therefore

<math>\frac{\left(x+h\right)^r - x^r}{h} = \sum_{k=1}^{r} {}^rC_k h^{k-1} x^{r-k}</math>

The derivative is the limit of this as <math> h \rightarrow 0 </math>

<math>\frac{d}{dx}\left(x^r\right) = \lim_{h\rightarrow 0} \left(\sum_{k=1}^{r} {}^rC_k h^{k-1} x^{r-k}\right) = {}^rC_1 x^{r-1} = rx^{r-1}</math>

which gives the claimed result.

Generalisation

<math>\frac{d}{dx} \left(ax^k\right) = akx^{k-1}</math>

is generally true for all values of k where x^k is meaningful. In particular it holds for all rational k for values of x where x^k is defined.

Similarly for integration, see Table of integrals.

References

Calculus of a Single Variable: Early Transcendental Functions (3rd Edition) by Edwards[?], Hostetler[?], and Larson[?] (2003)

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